Ruts gon BT A=3(2^2+1)(2^4+1)(2^8)(2^16+1)

0 bình luận về “Ruts gon BT A=3(2^2+1)(2^4+1)(2^8)(2^16+1)”

1. Đáp án:

    A=2^32-1

   Giải thích các bước giải:

   A=3(2^2+1)(2^4+1)(2^8+1)(2^16+1)

   A=3(2^2+1)(2^4+1)(2^8+1)(2^16+1)

      =(2^2-1).(2^2+1)(2^4+1)(2^8+1)(2^16+1)

      =(2^4-1).(2^4+1)(2^8+1)(2^16+1)

      =(2^8-1).(2^8+1)(2^16+1)

      =(2^16-1).(2^16+1)

      =2^32-1

   Trả lời